Poisson's equation and Green's functions

[jmb]

Trying to verify the general integral solution to Poisson's equation I reach the following contradiction, what am I missing/doing wrong?

A solution $$u(\mathbf{x})$$ to Poisson's equation satisfies:

$$\nabla^2 u(\mathbf{x}) = - \frac{\rho(\mathbf{x})}{\epsilon_0}$$

we can find such a solution in a given domain by evaluating:

$$u(\mathbf{x}) = \int_{V} \frac{\rho(\mathbf{x'})}{4 \pi \epsilon_0 |\mathbf{x}-\mathbf{x'}|} d\mathbf{x'}$$

or equivalently (if we are only interested in the electric field):

$$\mathbf{E} = -\nabla u = - \int_{V} \frac{\rho(\mathbf{x'})}{4 \pi \epsilon_0} \frac{\mathbf{x}-\mathbf{x'}}{|\mathbf{x}-\mathbf{x'}|^3} d\mathbf{x'}$$

We can think of these solutions as either the Green's function solution or, physically, as the convolution of the point source solution with the source density $$\rho(\mathbf{x})$$.

However if I take the Laplacian (or the divergence in the case of the electric field solution) of either of the above I get zero instead of $$-\frac{\rho}{\epsilon_0}$$. What am I missing??

To evaluate the Laplacian/divergence of the above I make use of the fact that $$\mathbf{x'}$$ is a 'dummy variable', which allows me to switch the order of integration and differentiation when applying $$\nabla^2$$ to the integral on the RHS and also means the components of $$\mathbf{x'}$$ all have derivative zero. With these assumptions the operation is fairly straightforward to carry out, but I have anyway verified my answer with Mathematica.

 

[marcusl]

Not so. You should take the Laplacian with respect to x (after all you are calculating $$\nabla^2{u(x)}$$). Since
$$\nabla^2 \frac{1}{|x-x'|}=\delta(x-x')$$
you get the right result from the integration.
This expression is derived in many places:
Jackson, Classical Electrodynamics, 2nd ed., Eq. (1.31)
Arfken, Math Methods for Physicists, 2nd ed., sect. 1.15
to give two examples.

 

[jmb]

Thanks marcusl.

Yes I was taking the derivative wrt x (hence the derivatives of all components of x' being zero). But I was erroneously evaluating the Laplacian of $$\frac{1}{|x-x'|}$$ as zero.

I've since found references to the result you quote, but they all start by searching for the Green's function $$G$$ that satisfies $$\nabla^2 G = \delta(x-x')$$ and then deducing it to be the above... out of interest do you know of any way to show it starting from the other end --- i.e. directly differentiating $$\frac{1}{|x-x'|}$$ and showing its Laplacian to be the delta function? Or do the references you quote do it that way?

I'll look them up when I get the chance. Thanks again.

 

[marcusl]

Yes, Jackson shows it explicitly. And, now that I'm looking at his formula instead of relying on memory, I see I forgot the constant
$$\nabla^2 \frac{1}{|x-x'|}=-4\pi \delta(x-x')$$

 

[samalkhaiat]

Yes, Jackson shows it explicitly. And, now that I'm looking at his formula instead of relying on memory, I see I forgot the constant
$$\nabla^2 \frac{1}{|x-x'|}=-4\pi \delta(x-x')$$

In my 2nd edition, Jackson says on p.40 :" since $\nabla^{2}(1/r) = 0$ for $r \neq 0$ and its volume integral is $-4\pi$, we can write the formal equation, $\nabla^{2}(1/r) = -4\pi \delta^{3}(x)$..."

In exam, if you reproduce "Jackson's proof", I give you 2 marks out of 10! You would gain the remaining 8 marks, if you prove the underlined sentence in Jackson's statement. I can tell you, it is a tricky one, and Jackson doesn't do it!

regards

sam

 

[jmb]

OK samalkhaiat I'll have a shot at your "remaining 8 marks", let me know if I'm missing something!

From the divergence theorem we have,

$$I = \int_{V} \nabla^2 (1/r) dV = \int_{S} \nabla (1/r) \cdot \mathbf{dS}$$

Since the RHS is a surface integral we don't need to worry about evaluating $$\nabla (1/r)$$ at the origin and so, choosing V as a sphere centred on the origin, we can straightforwardly write the surface integral as:

$$I = \int_{0}^{2\pi} \int_{0}^{\pi} \frac{\partial}{\partial r} \left(\frac{1}{r}\right) r^2 \sin{\theta} \; d\theta \; d\phi = -4\pi$$

(since $$\frac{\partial}{\partial r} \left(\frac{1}{r}\right) = -\frac{1}{r^2}$$) as required.

 

[marcusl]

Jackson has similar expressions on p. 39 (he calls the integral C instead of I).

 

[samalkhaiat]

OK samalkhaiat I'll have a shot at your "remaining 8 marks", let me know if I'm missing something!

From the divergence theorem we have,

$$I = \int_{V} \nabla^2 (1/r) dV = \int_{S} \nabla (1/r) \cdot \mathbf{dS}$$

Since the RHS is a surface integral we don't need to worry about evaluating $$\nabla (1/r)$$ at the origin and so, choosing V as a sphere centred on the origin, we can straightforwardly write the surface integral as:

$$I = \int_{0}^{2\pi} \int_{0}^{\pi} \frac{\partial}{\partial r} \left(\frac{1}{r}\right) r^2 \sin{\theta} \; d\theta \; d\phi = -4\pi$$

(since $$\frac{\partial}{\partial r} \left(\frac{1}{r}\right) = -\frac{1}{r^2}$$) as required.

For this, I give you 3 marks out of the 8! Your next step is to show that the result $-4\pi$ holds true for ANY surfase S bounding V (not just spherical). Then and only then you can earn the remaining marks! Have a go at it and let me know.


regards

sam

 

[jmb]

For this, I give you 3 marks out of the 8! Your next step is to show that the result $-4\pi$ holds true for ANY surfase S bounding V (not just spherical). Then and only then you can earn the remaining marks! Have a go at it and let me know.


regards

sam

Doesn't it just follow on from the properties of $$\nabla^2(1/r)$$?

We already know that $$\nabla^2(1/r)$$ is zero everywhere except $$r=0$$, thus any volume V (enclosed by some surface S) which encloses $$r=0$$ will produce the same value for the integral, regardless of its shape, since the integrand gives no contribution to the integral at any other location. Equally any volume not enclosing $$r=0$$ will give zero when integrated. Maybe I'm missing a subtlety?

 

[samalkhaiat]

We already know that $$\nabla^2(1/r)$$ is zero everywhere except $$r=0$$, thus any volume V (enclosed by some surface S) which encloses $$r=0$$ will produce the same value for the integral, regardless of its shape, since the integrand gives no contribution to the integral at any other location. Equally any volume not enclosing $$r=0$$ will give zero when integrated. Maybe I'm missing a subtlety?

By stating this, you gain the remaining marks. The trick is to surround the point x = 0 by small enough sphere and do the integration in the bounded region $x \neq 0$:

Let $\Sigma$ be a surface bounding a region V containing the orign, and let $\sigma$ be a small spherical surface enclosing the point x = 0. Therefore, we have $x \neq 0$ everywhere in the region U bounded by $\Sigma + \sigma$, i.e.,

$$\int_{U} \nabla^{2}|\frac{1}{x}| \ d^{3}x = 0 = \int_{\Sigma + \sigma} \vec{\nabla}|\frac{1}{x}| \ . \ d\vec{S}$$

From this it follows that the integral over the whole region V is

$$\int_{V} \nabla^{2}|\frac{1}{x}| \ d^{3}x = \int_{\Sigma} \vec{\nabla}|\frac{1}{x}| \ . \ d\vec{S} = - \int_{\sigma} \vec{\nabla}|\frac{1}{x}| \ . \ d\vec{S}$$

Or,

$$\int_{V} \nabla^{2}|\frac{1}{x}| \ d^{3}x = \int_{\sigma} \frac{1}{|x|^{2}} \left( \hat{x}. \hat{n}_{\sigma} \right) dS$$

where $\hat{x}$ is the unit vector along x and $\hat{n}_{\sigma}$ is the unit OUTWARD normal at dS of $\sigma$;

$$\hat{x} . \hat{n}_{\sigma} = -1$$

Thus

$$\int_{V} \nabla^{2}|1/x| \ d^{3}x = - \frac{1}{R^{2}} \int_{\sigma} dS = -4\pi$$


regards

sam

 

[jmb]

Cool. Many thanks again to both marcusl and samalkhaiat for your help!